This document provides details on calculating seismic loads according to ASCE 7-05, including: 1) Important notes on including both horizontal and vertical seismic effects in load combinations. 2) An overview of the equivalent lateral force procedure and its applicability. 3) Two examples that demonstrate applying the procedure to calculate seismic base shear for an ordinary steel concentric braced frame building and ordinary reinforced concrete shear wall building.

1. ASCE 7-05 SEISMIC LOAD CALCULATION
 Contents:
 Important notes on sisimic load effect on ASCE 7-05, Sec 12.4
 ASCE 7-05 Equivalent lateral force procedure
 Example 1: Building frame systems with ordinary steel concentric braced frame
 Example 2: Building frame systems with ordinary reinforced concrete shear wall
Important Note on seismic load effects on ASCE 7-05, Sec 12.4
1. Seismic design shall include both effective from horizontal seismic and vertical seismic force. The seismic force, E, appears in load
combinations
1.2 D 1.0 E + L*
+ 0.2S
0.9 D + 1.0 E
shall include both horizontal and vertical effects,
E = Eh Ev (AISC 7-05 Eq. 12.4.1 & 12.4.2)
where Eh is effect of horizontal seismic force, Ev is effect of vertical seismic force.
2. In seismic design category, D, E, & F (except special condition in section 12.3.4.1 and 12.3.4.2), horizontal seismic load effect shall
include redundancy factor, 1.3)
Eh = QE (AISC 7-05 Eq. 12.4.3.)
where QE is effect of horizontal seismic shear force calculated based on seismic base shear V, or Fp from equation for components,
non-structural elements, etc.

2. 3. For strength design, vertical seismic load effect shall be calculated as
Ev = 0.2 SDSD (AISC 7-05 Eq. 12.4.4)
where SDS is seismic parameter, D is effect of dead load.
The load combinations become
(1.2+0.2SDS) D QE + L*
+ 0.2S
(0.9- 0.2SDS) D + QE
4. For allowable stress design, vertical seismic load effect shall be include in load combination as
(1.0 + 0.14SDS)D + 0.7 QE
(1.0 + 0.105SDS)D + 0.525 QE + 0.75 L + 0.75 (Lr or S)
(0.6 - 0.14SDS)D + 0.7 QE
ASCE 7-05 Equivalent lateral force procedure
Applicability
Equivalent lateral force procedure is limited to be used in
1. Seismic design category A.
2. Seismic design category B and C, (except light framed construction, see 6)
3. Regular structure with T < 3.5 Ts in Seismic design category D, E, and F
4. Irregular structure in Seismic design category D, E, and F with horizontal irregularities Type, 2, 3, 4, and 5 and T < 3.5 Ts.

3. 5. Irregular structure in Seismic design category D, E, and F with vertical irregularities Type 4, 5a and 5b and T < 3.5 Ts .
6. Light framed construction in occupancy I & II, more than 3 stories high, or occupancy III & IV, more than 2 stories high, and
regular light frame structures with T < 3.5 Ts in seismic design caterogy D, E, and F.
where Ts = SD1/SDS.
Procedure
1. Determine weight of building, W.
2. Determine 0.2 second and 1 second response spectral acceleration, Ss and S1 from Figure 22-1 to 22-14
3. Determine Site class from Chapter 20 or from soil report.
4. Determine site coefficient, Fa, from Table 11.4.1.
5. Determine site coefficient, Fv, from Table 11.4.2
6. Determine adjusted maximum considered earthquake spectral response acceleration
parameters for short period, SMS and at 1 second period, SM1.
SMS = Fa Ss (Eq. 11.4-1)
SM1 = Fv S1 (Eq. 11.4.2)
7. Determine deign spectral response acceleration parameters for short period, SDS
and at 1 second period, SD1.
SDS =(2/3) SMS (Eq. 11.4.3)
SD1 =(2/3) SM1 (Eq. 11.4.4)
8. Determine Important factor, I, from Table 11.5.1
9. Determine Seismic design category from Table 11.6-1, & 11.6-2
10. Determine Response modification factor, R. from Table 12.2-1
check building height limitation
11. Determine seismic response coefficient from Eq. 12.8-1
Cs= SDS / (R/I)
12. Determine approximate fundamental period from Eq. 12.8-7
T = Ct hn
x
where
hn is the height of building above base.

4. Ct = 0.028, x = 0.8 for steel moment resisting frame,
Ct = 0.016, x = 0.9 for concrete moment resisting frame,
Ct = 0.03, x = 0.75 for eccentrically braced steel frame,
Ct = 0.02, x = 0.75 for all other structural frame
13. Check if T < 3.5 SD1/SDS.
14. Determine Maximum seismic response coefficient Eq. 12.8-3 and 12.8-4
Csmax= SD1 / [ T(R/I)] for T TL
Csmax= SD1 / [T2
(R/I)] for T TL
where TL is long-period transition period in Figure 22.15 to 22.20.
15. Minimum seismic response coefficient, Eq 12.8-5 & 12.8.6
Csmin 0.01
If S1 0.6g, Csmin 0.5 S1 / (R/I)
16. Determine Seismic response for strength design from 12.8-1
V = Cs W
Example 1: Building frame systems with ordinary steel concentric braced frame
Given:
Code: ASCE 7-05 Equivalent lateral force procedure
Design information:
Weight of building, W = 500 kips
0.2 second response spectral acceleration, Ss = 0.25
1 second response spectral acceleration, S1 = 0.1
Long-period transition period, TL = 12 sec
Building frame systems with ordinary steel concentric braced frame
Building category II
Building height: 30 ft
Soil Profile: E
Requirement: Determine seismic base shear

5. Solution:
Site coefficient, Fa = 2.5
Site coefficient, Fv = 3.5
Design spectral response acceleration parameters
SMS = Fa Ss = 0.625 (Eq. 11.4.1)
SM1 = Fv S1= 0.35 (Eq. 11.4.2)
SDS =(2/3) SMS =0.417 (Eq.11.4.3)
SD1 =(2/3) SM1 =0.233 (Eq. 11.4.4)
Seismic design category B from Table 11.6.1, category C based on Table 11.6.2.
Use category D.
From Table 12.2.1, No limit for building frame system with ordinary steel concentric braced frame
Response modification factor, R = 3.25
Important factor, I = 1 (Table 11.5.1)
Seismic response coefficient (Eq. 12.8-1)
Cs= SDS / (R/I) = 0.128
Fundamental period (Eq. 12.8-7), CT = 0.02, x = 0.75
T = CT hn
0.75
= 0.256 sec
< (3.5)(SDS/SD1) = (3.5)(0.417/0.233)=1.96 O.K.
Maximum seismic response coefficient (Eq. 12.8-3)
T < TL = 12 sec
Csmax= SD1 / [(R/I) T] = 0.192
Since S1 = 0.1 < 0.6, Minimum seismic response coefficient (Eq. 12.8-5)
Csmin= 0.01
Seismic base shear for strength design
V = Cs W = (0.128)(500) = 64 kips
Example 2: Building frame systems with ordinary reinforced concrete shear wall
Given:
Code: ASCE 7-05 Equivalent lateral force procedure
Design information:
Weight of building, W = 1000 kips

6. 0.2 second response spectral acceleration, Ss = 0.5
1 second response spectral acceleration, S1 = 0.15
Soil profile class: C
Building frame systems with ordinary reinforced concrete shear wall
Building category II
Building height: 40 ft
Requirement: Determine seismic base shear for strength design
Solution:
Site coefficient, Fa = 1.2
Site coefficient, Fv = 1.65
Design spectral response acceleration parameters
SMS = Fa Ss = 0.6 (Eq. 11.4.1)
SM1 = Fv S1= 0.247 (Eq. 11.4.2)
SDS =(2/3) SMS =0.4 (Eq. 11.4.3)
SD1 =(2/3) SM1 =0.165 (Eq. 11.4.4)
Seismic design category C from Table 11.6.1, category C based on Table 11.6.2.
Use category C.
From Table 12.2.1, no limit for building frame system with ordinary shear wall for category C.
Response modification factor, R = 5
Important factor, I = 1 (Table 11.5.1)
Seismic response coefficient (Eq. 12.8-1)
Cs= SDS / (R/I) = 0.08
Fundamental period (Eq. 12.8-7), CT = 0.02, x = 0.75
T = CT hn
0.75
= 0.318 sec
< (3.5)(SDS/SD1) = (3.5)(0.4/0.165)=1.44 sec O.K.
Maximum seismic response coefficient (Eq. 12.8-3)
T < TL = 12 sec
Csmax= SD1 / [(R/I) T] = 0.104
Since S1 = 0.15 < 0.6, Minimum seismic response coefficient (Eq. 12.8-5)
Csmin= 0.01
Seismic base shear for strength design
V = Cs W = (0.08)(1000) = 80 kips